A box of books weighing 350 N is shoved across the floor of an apartment by a force of 460 N exerted downward at an angle of 35.0° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 3.80 m, starting from rest?

200

Well, let me calculate it for you. But first, let me ask you this: why did the box refuse to tell the time? Because it didn't want to get ticked off! Now, let's get to the calculation.

To find out how long it takes to move the box 3.80 m, we need to use some physics formulas. The first one we'll use is the horizontal component of the force, which is given by:

F_horizontal = F_applied * cos(angle)

Substituting the given values, we have:

F_horizontal = 460 N * cos(35.0°)

Next, we calculate the frictional force using the formula:

F_friction = coefficient * normal force

The normal force is the weight of the box, which is equal to its mass multiplied by gravity:

F_normal = m * g

Now, we can calculate the time it takes to move the box using the formula:

time = distance / velocity

To find the velocity, we use the equation:

acceleration = (F_horizontal - F_friction) / mass

Now, let me plug in the numbers and crunch the calculations for you. Let's clown it up and solve this!

Calculating the horizontal component of the force:
F_horizontal = 460 N * cos(35.0°)

Calculating the normal force:
F_normal = 350 N

Calculating the frictional force:
F_friction = 0.57 * F_normal

Calculating the acceleration:
acceleration = (F_horizontal - F_friction) / mass

Calculating the velocity:
velocity = sqrt(2 * acceleration * distance)

Calculating the time:
time = distance / velocity

Unfortunately, I'm just a bot and I don't have all the information needed to calculate these values accurately. It seems like we need the mass of the box. So, without that information, I can't give you a specific answer for how long it takes to move the box 3.80 m. However, I hope my attempted humor lightened up the situation a bit!

To find the time it takes to move the box, we need to break down the forces acting on the box and calculate the acceleration.

1. Resolve the force of 460 N into horizontal and vertical components:
- The horizontal component is 460 N * cos(35.0°) = 460 N * 0.819 = 376.14 N.
- The vertical component is 460 N * sin(35.0°) = 460 N * 0.573 = 263.88 N.

2. Calculate the net horizontal force:
- The net horizontal force is given by F_net = F_applied - F_friction.
- F_applied is the horizontal component of the applied force = 376.14 N.
- F_friction is the force due to friction = μ * F_normal, where μ is the coefficient of kinetic friction.
- F_normal is the normal force acting on the box, which is equal to the weight of the box = 350 N.
- F_friction = 0.57 * 350 N = 199.5 N.
- F_net = 376.14 N - 199.5 N = 176.64 N.

3. Calculate the acceleration:
- The net force is related to acceleration by Newton's second law: F_net = m * a, where m is the mass of the box.
- The weight of the box is given by the equation: weight = m * g, where g is the acceleration due to gravity.
- Solving for mass, we have: m = weight / g = 350 N / 9.8 m/s^2 = 35.71 kg.
- Therefore, F_net = m * a = 35.71 kg * a = 176.64 N.
- Solving for acceleration, we have: a = F_net / m = 176.64 N / 35.71 kg ≈ 4.94 m/s^2.

4. Use the kinematic equation to find the time:
- The kinematic equation we can use is: d = v_i * t + (1/2) * a * t^2, where d is the displacement, v_i is the initial velocity, and t is the time.
- The initial velocity is zero since the box starts from rest.
- Plugging in the values, we have: 3.80 m = 0 * t + (1/2) * 4.94 m/s^2 * t^2.
- Simplifying, we get: 3.80 m = (1/2) * 4.94 m/s^2 * t^2.
- Rearranging the equation, we have: t^2 = (2 * 3.80 m) / 4.94 m/s^2.
- Solving for t, we get: t ≈ √((2 * 3.80 m) / 4.94 m/s^2) ≈ √(1.54 s^2) ≈ 1.24 s.

Therefore, it takes approximately 1.24 seconds to move the box 3.80 meters, starting from rest.

To determine the time it takes to move the box, we can break down the problem into several steps:

Step 1: Determine the vertical and horizontal components of the force applied.

First, let's determine the vertical component of the force:

Vertical force = force applied * sin(angle)
Vertical force = 460 N * sin(35.0°)
Vertical force = 460 N * 0.5745
Vertical force ≈ 264.57 N

Next, let's determine the horizontal component of the force:

Horizontal force = force applied * cos(angle)
Horizontal force = 460 N * cos(35.0°)
Horizontal force = 460 N * 0.8192
Horizontal force ≈ 376.19 N

Step 2: Calculate the force of friction.

Friction force = coefficient of kinetic friction * normal force
Normal force = weight of the box (mg)
Normal force = mass * acceleration due to gravity
Normal force = weight of the box / gravitational acceleration
Normal force = 350 N / 9.8 m/s^2
Normal force ≈ 35.71 kg

Friction force = 0.57 * 35.71 kg * 9.8 m/s^2
Friction force ≈ 197.02 N

Step 3: Calculate the net force.

The net force is the horizontal component of the applied force minus the friction force:

Net force = horizontal force - friction force
Net force = 376.19 N - 197.02 N
Net force ≈ 179.17 N

Step 4: Calculate the acceleration.

Using Newton's second law, we can determine the acceleration:

Net force = mass * acceleration
Acceleration = net force / mass
Acceleration = 179.17 N / 35.71 kg
Acceleration ≈ 5.02 m/s^2

Step 5: Use the kinematic equation to calculate the time.

We'll use the kinematic equation that relates displacement, initial velocity, time, and acceleration:

Displacement = (initial velocity * time) + (0.5 * acceleration * time^2)

Since the box starts from rest (initial velocity = 0), the equation simplifies to:

Displacement = 0.5 * acceleration * time^2

Rearranging the equation, we can solve for time:

time^2 = (2 * displacement) / acceleration
time^2 = (2 * 3.80 m) / 5.02 m/s^2
time^2 ≈ 1.51 s^2

Taking the square root of both sides, we find:

time ≈ √(1.51 s^2)
time ≈ 1.23 s

Therefore, it takes approximately 1.23 seconds to move the box a distance of 3.80 meters, starting from rest.